Questions: Full and Faithful Functors

Question 1

F: C → D is a fully faithful functor and F(f): F(A) → F(B) is an isomorphism in D. What can you conclude about f in C?

• A. Nothing — F may create isomorphisms that do not exist in C
• B. f must be an isomorphism in C, because fully faithful functors reflect isomorphisms
• C. F must be surjective on objects, because the isomorphism must come from C
• D. f must be an identity morphism

Answer: (B)

A fully faithful functor reflects isomorphisms: if F(f) is an isomorphism in D, then f was already an isomorphism in C. This follows because F is bijective on each hom-set — the inverse of F(f) must be in the image of F, so there exists g: B → A in C with F(g) = F(f)⁻¹, and faithfulness forces f and g to be inverses. Option A describes a functor that is merely full or faithful but not both. Option C confuses object-surjectivity with the reflection property.

Question 2

The forgetful functor U: Grp → Set sends each group to its underlying set and each group homomorphism to the same function between sets. Which properties does U have?

• A. Full but not faithful, because not every set function is a group homomorphism
• B. Faithful but not full, because distinct group homomorphisms yield distinct set functions, but not every set function between groups is a homomorphism
• C. Fully faithful, because it preserves all morphism information
• D. Neither full nor faithful, because groups have more structure than sets

Answer: (B)

Faithful: if two group homomorphisms f, g: G → H are equal as set functions, they are equal as homomorphisms — so U is injective on hom-sets. Not full: there exist set functions between groups that are not group homomorphisms (e.g., most constant maps). Fullness would require every set function to arise from a group homomorphism, which fails. This is the prototypical example of a faithful-but-not-full functor.

Question 3

A fully faithful functor F: C → D should be surjective on objects — most object in D is in the image of F.

• T. True
• F. False

Answer: False

Fully faithful only describes how F behaves on morphisms between pairs of objects (bijective on each hom-set). It says nothing about whether every object in D is hit. The Yoneda embedding is fully faithful but embeds a small category into a much larger presheaf category with far more objects. The failure of surjectivity on objects is exactly why a fully faithful functor need not be an equivalence of categories.

Question 4

A functor that is faithful should be injective on objects — if F(A) = F(B) then A = B.

• T. True
• F. False

Answer: False

Faithful means injective on each hom-set: F(f) = F(g) implies f = g for parallel morphisms f, g: A → B. It says nothing about objects. A faithful functor can send many distinct objects to a single object in D — it just cannot confuse two morphisms between the same pair of objects. The hom-set condition and the object condition are entirely independent.

Question 5

Explain the difference between a functor being 'full' and being 'faithful,' and give an example of a functor that is one but not the other.

Model answer: Faithful means the functor is injective on each hom-set: distinct morphisms f ≠ g in C(A,B) map to distinct morphisms F(f) ≠ F(g) in D(FA,FB). Full means surjective on each hom-set: every morphism in D(FA,FB) arises as F(f) for some f in C(A,B). The forgetful functor Grp → Set is faithful (same set-function implies same homomorphism) but not full (not every set function is a group homomorphism). The inclusion of a non-full subcategory is faithful but not full, because there exist morphisms between the included objects in the larger category that are absent from the subcategory.

A faithful functor loses no information about morphisms between given objects. A full functor has no 'extra' morphisms between images beyond what came from C. Fully faithful means D(FA,FB) and C(A,B) are in bijection for every pair A,B — making F an embedding of morphism structure, though not necessarily of objects.